Question

After how many second will the concentration of the reactant in a first order reaction be halved if the rate constant is $1.155\times {10}^{-3}{s}^{-1}$?

• A. 600
• B. 100
• C. 60
• D. 10

Answer 1

The correct option is A 600

Rate constant $k=1.155\times {10}^{-3}{s}^{-1}$
$k=\frac{2.303}{t}\log \frac{a}{(a-x)}\because a=a,(a-x)=\frac{a}{2}$
$t_{1/2}=\frac{2.303}{k}\log \frac{a}{a/2}$
$=\frac{2.303}{1.155\times {10}^{-3}}\log 2$
$=\frac{2.303}{1.155\times {10}^{-3}}\times 0.3010$
$=\frac{0.693\times {10}^3}{1.155}$
or $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.155\times {10}^{-3}}$
$=600s$